# Tag Info

22

You're actually hitting on a very famous concept here that revolutionized physics!! Your understanding is almost wholly correct and your analogy is a good one - excellent reasoning - the only thing missing is radiation from the system. This latter lack is mostly irrelevant for the level of question you have been thinking about: but I'll address that below. ...

18

Nerd Sniping! The answer is $\frac{4}{\pi} - \frac{1}{2}$. Simple explanation: http://www.mbeckler.org/resistor_grid/ Mathematical derivation: http://www.mathpages.com/home/kmath668/kmath668.htm

17

First, Field strength. This calculation is strictly an electric potential calculation; radiation and induction are safely ignored at 50Hz.* For a 200kV transmission line 20m above ground, the max electric field at ground level is about 1.2 kV/m. This number is reduced from the naive 200kV/20m=10 kV/m calculation by two effects: 1) The ~1/r variation in ...

13

It's not true. To see this, you can try an experiment with some batteries and light bulbs. Hook up two bulbs of different wattages (that is, with different resistances) in parallel with a single battery: ------------------------------------------ | | | Battery Bulb 1 Bulb 2 | ...

13

where does that electricity go? The photons from the sun have energy and momentum, but not "electricity". Essentially, a photon (solar or otherwise) striking the solar panel can create an electron-hole pair (EHP) and, if the EHP is within or near the depletion zone, the pair will be separated by the built-in electric field. This results in a ...

12

$\def\vE{{\vec{E}}}$ $\def\vD{{\vec{D}}}$ $\def\vB{{\vec{B}}}$ $\def\vJ{{\vec{J}}}$ $\def\vr{{\vec{r}}}$ $\def\vA{{\vec{A}}}$ $\def\vH{{\vec{H}}}$ $\def\ddt{\frac{d}{dt}}$ $\def\rot{\operatorname{rot}}$ $\def\div{\operatorname{div}}$ $\def\grad{\operatorname{grad}}$ $\def\rmC{{\mathrm{C}}}$ $\def\rmM{{\mathrm{M}}}$ $\def\ph{{\varphi}}$ ...

11

Yes, it is possible. For example Kevin Brown did here and here including this table. so for the xkcd problem the answer is $-\frac{1}{2}+\frac{4}{\pi} \approx 0.773$.

10

Yes Sam, there definitely is electric field reshaping in the wire. Strangely, it is not talked about in hardly any physics texts, but there are surface charge accumulations along the wire which maintain the electric field in the direction of the wire. (Note: it is a surface charge distribution since any extra charge on a conductor will reside on the ...

10

Electric current, by definition, is a flow of charged particles. When someone says it is the propagation of the electric field, usually he means the following: The velocity of the electrons in the wires is very slow (few cm/s if I remember it right), but when one turn on the light he doesn't see any delay. The lamp starts lighting when the electrons start ...

10

AC or DC, you only get electrocuted if current passes through your body. (Current passing through any part of your body can be dangerous, and possibly cause an electrical burn, but current passing across your heart is the one that's really dangerous.) Touching just one wire at a time gives the current nowhere much to go. You are right to think that some ...

10

Batteries do not behave in such an ideal way across all conditions. The simplest model of a battery as a circuit element is the one you describe - a pure voltage source. A slightly-more sophisticated model is as a voltage source connected to a fixed resistor, called the battery's internal resistance. A typical battery has an internal resistance of between 1 ...

10

instead of thinking your body is empty and that a charged wire has to push electrons one by one through you and into the ground (blood is actually full of charge carriers), a better analogy would be a very long queue of pushy people. if the entrance to the apple store doesn't open, it doesn't matter how hard the guy at the back pushes--nothing moves. ...

9

If the power line is 20m high, and has the voltage of 1MV , then the electric field (near ground), very roughly, is on order of 1000/30 kv ~ 30 000 v/m (the numbers are very approximate and the field is complicated because it is a wire near a plate scenario, and wire diameter is unknown but not too small else the air would break down, i.e. spark over, near ...

9

A human body may reflect and absorb radio frequencies, though not very efficiently. It may as well act as a resonance chamber for certain frequencies. For a signal of 100 MHz, the involved wavelength is 3 m, and so it is possible that parts of your body are acting slightly as a resonant chamber. (for an optimal resonance, you should have 1.5 m diameter, too ...

8

Why are wires in simple circuits approximated as equipotentials? Because one of the three assumptions of circuit theory is: All electrical effects happen instantaneously throughout a circuit. If the circuit is small enough compared to the wave length of the signals applied, all electric signals travel through it so quickly, that we can assume that they ...

8

I find this sort of thing becomes much more intuitive if you can think of an analogy in terms of water. In this case, we can think of it like this: Here we have water flowing through a hole in a bath tub, into another tub underneath. The stick figure has been given the task of keeping the water level constant, by lifting water back up into the top tub ...

7

First, your camera is not designed to work with batteries below a certain voltage. When it detects an excessively low battery voltage it turns itself off. That circuit stays in the "off" state until voltage is completely removed from the circuit. When you operate your camera, the current required by your camera varies according to what you do with it. So ...

7

Negative current just means it flows opposite to the chosen direction. Take your example, you have a loop with increasing area and downward pointing B-field x x x x x x x x x +-------------|----------- ^ x | x x x x | x x x | y | | ---> h ^ x | x x x x | x x x | | ...

7

Virtual ground refers to a circuit element not directly connected to ground, held at a reference voltage. This reference voltage need not be the same voltage as ground either. For example many op-amp circuits were originally designed for dual power supplies (say +12V and -12V) and could handle filtering or modification of a signal that was oscillating ...

7

"Ground" refers to a particular voltage, generally taken to be "zero", or the voltage of the earth. A "virtual ground" is a wire in a circuit whose voltage is held to be zero not because it is directly connected to the true ground, but instead because it is actively driven to that voltage typically by feedback mechanisms. Here's an example of a virtual ...

7

I'll take it step by step here. First I'll write the answer for the first few cases with circuit analysis. Then I'll apply a reduction to show the pattern that the problem arrives at. N=1 $$Z = R+R=2R$$ N=2 $$Z = R+\frac{1}{\frac{1}{R}+\frac{1}{R + R}} = R \left( 1+\frac{1}{1+\frac{1}{1 + 1}} \right)=\frac{5}{3} R$$ N=3 $$Z = ... 7 There's an old and clichéd but actually pretty good analogy for understanding electric circuits, and that's to think of the circuit as water plowing through pipes. If you have a pipe full of water, and you turn on the tap at one end, water immediately starts flowing out of the other end. Well not quite immediately: when you turn on the tap you raise the ... 7 I'll give the answer to this question using an unusual method that showed up in the American Mathematical Monthly's problem section perhaps in the late 1970s. This is not necessarily the easy way to solve the problem, but it works out nicely from an algebraic point of view. The way most people solve most resistance problems is to use series and parallel ... 7 The three capacitors are connected in parallel. There are only two nodes in this circuit. A series connection requires at least three. The equivalent capacitance is just the sum of the three capacitances. UPDATE: The circuit can be redrawn such that the parallel connection is manifest. 7 Electrons that reach the positive terminal indeed remain there. The potential difference between the two terminals pushes electrons from the negative anode toward the positive cathode. When an electron reaches the cathode, it stays there to equalize the original charge imbalance between the two nodes. When electrochemical redox reaction sustaining the ... 7 Actually, induction works, although it is often used a bit differently than you described. You can place a warm superconductor loop into a normal coil. As you switch the coil on, there will be some current inside the superconductor, but since it is not cold yet, this current quickly dies down. Then you cool the superconductor below its critical temperature. ... 7 Indeed, AC can flow without a "complete circuit" - that's what happens in LC circuits all the time. An LC circuit is technically not complete - the capacitor of LC circuit contains an insulator between its plates and so electrons are unable to flow through the capacitor (unless it fails). Still the oscillations in the LC circuit happen because of alternating ... 7 Alfred got in before me, but I have a diagram! I've marked all continuous bits of wire in the same colour, and marked the corresponding colours on the ends of the resistors. A quick redraw later and I get: which is a lot simpler! 7 Gregsan's and Kieran's answers are insightful analogies and the pushy electrons are certainly part of the answer. There is another aspect to the "decision" process and that is the propagation of electromagnetic waves. There is a chapter in the second volume of the Feynman Lectures on Physics - I don't have it with me but the relevant section will be just ... 7 If you have just given the voltage signal with$$ \def\l{\left}\def\r{\right} v(t) = \l(2-\l|\frac t{\rm s}-2\r|\r)\rm V $$then the current at t=2\rm s is undefined. Right. But, in most cases really nobody cares. What we learn theoretically about the current from the above voltage signal definition is that$$ i(t) = \begin{cases} C\cdot 1\frac{\rm V}{\rm ...

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